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Metamath Proof Explorer

Theorem eldisjssd

Description: Subclass theorem for disjoint elementhood, deduction version. (Contributed by Peter Mazsa, 28-Sep-2021)

		Ref	Expression
	Hypothesis	eldisjssd.1	$⊢ φ \to A \subseteq B$
	Assertion	eldisjssd	$⊢ φ \to (ElDisj B \to ElDisj A)$

Step	Hyp	Ref	Expression
1		eldisjssd.1	$⊢ φ \to A \subseteq B$
2		eldisjss	$⊢ A \subseteq B \to (ElDisj B \to ElDisj A)$
3	1 2	syl	$⊢ φ \to (ElDisj B \to ElDisj A)$